In my harmonic analysis class I have to prove that for all $a>0$ the following equality holds: $$\sum_{n\in \mathbb{Z}}e^{-n^2\pi a}=\frac{1}{\sqrt{a}}\sum_{k\in \mathbb{Z}}e^{-k^2 \pi / a}.$$

I'd appreciate any help on how to approach this problem and where to start.

Thank you!


I tried to apply Poisson's summation formula. I guessed that the function is $$f(x)=\exp(-\frac{x^2a}{4\pi}).$$ On the right I get $\sum_{n\in \mathbb{Z}}e^{-n^2\pi a}$, as required. But for the left side I have to calculate the integral $$\int_{-\infty}^{\infty}f(x)\exp(-i\omega x) dx ,$$ and I can't find the right substitution to solve it.

Any help would be appreciated!


I came to the answer (after the useful hints), so in order not to leave this question open, I will answer it myself.

The main idea is to use Poisson summation formula: $$\sum_{k\in \mathbb Z} f(2\pi k) = \frac{1}{2\pi} \sum_{n\in \mathbb Z} \hat{f}(n). $$ We start with setting $f(x)=e^{-\frac{x^2}{4\pi a}}$, and thus $f(2\pi k)= e^{-\frac{k^2\pi}{a}}.$ Now we have to find $\hat{f}(\omega)$. At first I thought to solve the above-mentioned integral, but its not that easy. So I took a different approach (thought I couldn't remember where I've seen it before): we can differentiate $f(x)$ to get a differential equation $$f'(x)+\frac{x}{2\pi a} f(x)=0.$$ Applying Fourier transform to this equation we get $$\omega \hat{f}(\omega)+\frac{1}{2\pi a} (\hat{f})'(\omega).$$ Solving this diff. equation we get $\hat{f}(\omega)=2\pi\sqrt{a}e^{-\pi a \omega^2}$ and that proves the claim.


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